Implicit Differentiation Question

[Tribo]

<< Mentor Note -- thread moved from the technical math forums at OP request, so no Homework Help Template is shown >>

x²y + xy² = 6

I know we use the chain rule from here, so wouldn't that be:

(d/dx)(x²y + xy²) = (d/dx)(6)

so using the chain rule of g'(x)f'(g(x) and the d/dx canceling out on the left side ...

2x + (d/dy)(4y²) = 0

Subtract 2x to the other side and cancel out the 4y² and I got -x/2y², which isn't right at all. :/

 

[Tribo]

...And I just saw in the rules I wasn't supposed to post this here. My bad, is there anyway to move it?

 

[fresh_42]

x²y + xy² = 6

I know we use the chain rule from here, so wouldn't that be:

(d/dx)(x²y + xy²) = (d/dx)(6)

so using the chain rule of g'(x)f'(g(x) and the d/dx canceling out on the left side ...

2x + (d/dy)(4y²) = 0

Subtract 2x to the other side and cancel out the 4y² and I got -x/2y², which isn't right at all. :/

First thing: I assume, that $y$ isn't dependent on $x$. You should mention this. Next, what happens to a factor when you differentiate?

 

[Tribo]

First thing: I assume, that $y$ isn't dependent on $x$.

It says, "use implicit differentiation to find dy/dx," so sure?

Next, what happens to a factor when you differentiate?

It becomes a derivative of the original?

 

[fresh_42]

It says, "use implicit differentiation to find dy/dx," so sure?

Where did it say this?

It becomes a derivative of the original?

Anyway, if $y=y(x)$ then you have to use the product rule and the chain rule.

Edit: I assume you meant the theorem of implicit functions.

 

[Tribo]

Where did it say this?

That's all the instructions said, although maybe the pages before had more information (I tired to read them but they didn't make any sense so I came here). To be frank I don't know why saying y isn't dependent on x is important or even quite what that means.

Edit: I assume you meant the theorem of implicit functions.

The explanatory blue box doesn't mention theorems but it does say this:

Implicit Differentiation
1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x.
2. Collect the terms with dy/dx on one side of the equation and solve for dy/dx.

So I guess it's saying y IS dependent on x? Or they have a relationship? Which means it's a function, if that means anything?

I don't know, this is really hard!

 

[Tribo]

I'll try it again with the product rule...

 

[Ray Vickson]

That's all the instructions said, although maybe the pages before had more information (I tired to read them but they didn't make any sense so I came here). To be frank I don't know why saying y isn't dependent on x is important or even quite what that means.
The explanatory blue box doesn't mention theorems but it does say this:

Implicit Differentiation
1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x.
2. Collect the terms with dy/dx on one side of the equation and solve for dy/dx.

So I guess it's saying y IS dependent on x? Or they have a relationship? Which means it's a function, if that means anything?

I don't know, this is really hard!

You start with an equation involving both $x$ and $y$, so for any given value of $x$ you can solve the equation to get one or more values of $y$. In your case, when you give $x$ a value such as $x = 1$ or $x = -7$ (or whatever) you will get a quadratic equation in $y$ that has two roots that you can find by the elementary methods you studied years ago.

So, indeed: we are interested in the case where y depends on x---very much so! The only problem is that although y depends on x we do not have an actual formula that tells us exactly what that dependence is. That is, while we know that we will have $y = g(x)$ for some function $g$ (that is, the function $g(x)$ exists), we don't have a formula for $g(x)$. The question is asking you how you would find the derivative $dy/dx= dg(x)/dx$ without actually knowing the formula for $g$ in the relationship $y = g(x)$.

 

[Tribo]

You start with an equation involving both $x$ and $y$, so for any given value of $x$ you can solve the equation to get one or more values of $y$.

Okay, just to make sure I'm understanding:

x + y = 6
y = 6 - x
So if x = 2, 4, 5, that would make y = 4, 2, 1, and so on.

In your case, when you give $x$ a value such as $x = 1$ or $x = -7$ (or whatever) you will get a quadratic equation in $y$ that has two roots that you can find by the elementary methods you studied years ago.

Hmm, okay. I don't really see it?

x²y + xy² = 6

Using the product rule I got:

(2xy + x2) + (y2 + 2y) = 0

The parentheses might be wrong, and I might've taken the derivative of 6 too soon, but I don't see where to go from here regardless. Surely it's not quadratic or going to be if there's no longer anything squared? Also... how do I chain 2xy?

So, indeed: we are interested in the case where y depends on x---very much so! The only problem is that although y depends on x we do not have an actual formula that tells us exactly what that dependence is. That is, while we know that we will have $y = g(x)$ for some function $g$ (that is, the function $g(x)$ exists), we don't have a formula for $g(x)$. The question is asking you how you would find the derivative $dy/dx= dg(x)/dx$ without actually knowing the formula for $g$ in the relationship $y = g(x)$.

What you're saying makes sense but I don't see it in the problem. We have y = a function but we don't know what the function looks like (it's formula).

On a side note, dy/dx = dg(x)/dx means
the derivative of y in respect to the derivative of x = the derivative of function g(x) in respect to the same derivative of x

...but what does that MEAN?

 

[SammyS]

...

Hmm, okay. I don't really see it?

Using the product rule I got:

(2xy + x2) + (y2 + 2y) = 0

That's incorrect.

Looking at the derivative of your first term:

The derivative (w.r.t. x) of x²y is:

(the derivative of x² ) times ( y )
PLUS
(x² ) times ( the derivative if y )

You have the first part right, but in the second part, you're missing the derivative of y .